Homeomorphism and the group structure on a circle | Algebraic Topology 2 | NJ Wildberger

[@Hbdhhkkj]

Finally a good camera man! BRAVO, can you have him instruct other camera people?

[@baptisteferrand8074]

Very didactic, rich in examples, clear and interesting to follow (which is not so common among math teachers). Thank you for sharing sir.

[@hanancohen8637]

Spectacular classes, very, very nice job. Please, let the classes going on. Greetings from Brazil.

[@solidstatejake]

Thank you for making your wonderful lectures available, Dr. Wildberger.

[@water0heaven]

For the latter two problems, I get the group multiplication rule as alpha(x1)*alpha(x2)=alpha(x1+x2) for the parabola and beta(x1)*beta(x2)=beta(x1x2) for the hyperbola, the fitness of which are not difficult to check. Together with the case of the circle, all the apparently distinct formulas seem to result from the same geometrical interpretation as drawing the parallel intersects, which is really amazing.

[@Igdrazil]

Excellent. A similar group structure works on a Elliptic Curve giving the most efficient cryptographic protection needed for small devices like smartphones etc that can't compute to fast but need nevertheless to be secured. A kind of "asymmetry" pin's out in your excellent presentation. Indeed through many aspects, Projective Geometry shows to be an unifying geometry. But that makes the "line at infinity" and thus "infinity" a central concept. A "miror" of the CENTER. And thus in this regard A and S1 are HOMEOMORPHIC because in projective sens, both ends are well connected...

[@francescoapg]

thanks for sharing!

[@JustNow42]

Wow. Excellent.

[@RichardSouthwell]

Its very good to present math like this on youtube, its great, it makes other people want to upload their mathematics on to KZbin too

[@allehelgen]

well, you do it well too !

[@dkwc4019]

wonderful presentation!! thanks alots !

[@njwildberger]

Hi, I use a HD Sony Handicam. But I am pretty sure any HD videocam would do just fine; it might be a good idea also to invest in a lapel wireless microphone, which helps improve the sound quality. Some good lighting to remove shadows is also something to think about. Good luck!

[@brendawilliams8062]

Thankyou. Dr. Wuldberger

[@pajeetsingh]

Your playlist is listed so down in search.

[@SatishJadhav-rr1oq]

It's really interesting ...and the way you teach the subject is very nice And nice to here 16:30 Indian name...(highschool teacher).

[@durgaaratikatla8729]

Really nice explanation 👍👌👌👌

[@maartenvanhoven1096]

Dr Wildberger, I am thoroughly enjoying your lectures! Thank you. With regard to the discussed group structure on the circle, I might be missing the point. Isn’t it trivial once we see that the product is actually the sum of circular arcs: A*B=C means arc(OA)+arc(OB)=arc(OC)? The associative rule is particularly trivial. Am I missing something?

[@TomPps]

Thank you so much for your videos they're really helpful

[@scottvska]

Thanks for uploading these.

[@mohdibrahimsheikh5181]

circle group The group structure on a circle is the work of F. Lemmermeyer, S. Shirali found in the Mathematical Gazette, Vol 93 #526, March 2009. Given points A, B, and C on the circle their products are also found on the circle giving the group property of closure. Group properties are the following: 1) Closure property as mentioned above.2) The point O acts as the group identity such that O*A=A*O=A.3) For every point A there is a point B such that A*B=B*A=O existence of inverse.4) (A*B)*C=A*(B*C), associative law.

[@sguzzygang]

Good vid!

At 47:57, what about the case h_1=1/h_2? Anyway, after algebrizing the geometrical group structure myself with basic Cartesian geometry I've got that h_3(h_1,h_2)=|(1-h_1h_2)/(h_1+h+2)| if h_1+h_2 eq 0 and h_3(h_1,h_2)=0 if h_1+h_2=0. And, with this, I think that the correct approach to define algebrically this circle group would be to firstly define your function ''e'', take its image as being the set/realm of the group and on this, define the operation A*B=e(h_3(e^{-1} (A) ,e^{-1} (B))), show that this definition is well defined (which is equivalent with showing ''e'' is bijective), and verify the group laws. EDIT: this would be a great setup if we would have a proper theory of functions of realms.

[@maximuspower2]

Wow very nice presentation of the group structure on the unit circle. However, I think it's the same as the regular group structure: namely, if e^{i\theta_1}, e^{i\theta_2} are on the circle, that group structure will have the same multiplication, e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2}.

[@Mikeecco]

I would love to check my thoughts on the final few problems. Is there a posting of the answers anywhere?

[@brendawilliams8062]

Thankyou so much.

[@connorfrankston5548]

Thank you Prof. Wildberger for all of your great and very interesting videos! I'm going through these lectures now, and I'm wondering what your opinion about category theory is. You might touch on that in this series but I'm not sure at this point. And on a totally unrelated note, I wonder what your opinion of geometric algebra and geometric calculus is. Finally, an alternative approach to the group structure of a circle could use rotation matrices expressed in terms of the rational parametrization of the circle. And now I see that is exactly what you used for Problem 4!

[@njwildberger]

Hi Connor, I think category theory is a useful meta language in which to discuss large scale mathematical structures, but it does not yet hold as a logical theory within maths on account of the set theoretical ambiguities. As for geometric algebra, that is also tricky. I will eventually discuss that.

[@carolinafraga7872]

I wish I had a teacher like you

[@jiguodai3533]

The problem 4 at 48:13, shouldn't be $h3 = \frac{h1*h2-1}{h1+h2}$ instead of $h3 = \frac{h1+h2}{1-h1*h2}$ ?

[@callumscott5107]

Fogive me if this is covered later on in the video but I've just learned that apparently my intuitions about homeomorphisms are mistaken and yet I feel fairly confident in my reasoning. I'm particularly interested in the dinstinction between the topological spaces and the symbols we attach to such spaces because I think it's of vital importance to do so. In a prior video, you discussed the idea of an ant travelling from A -> B, and then when it gets to B we move it onto a separate interval of B -> A, and you said that such a space could be simplified as simply A->A, but I feel that these are completely different because one only has a single distinct point of reference and the other has two. When you discuss the idea of an interval [0,1) earlier on in this video, you say that it isn't homeomorphic to a circle because 0 and 1 would map to the same point, but again I feel like the devil is in the details because suppose that you ONLY identified the points 0 and 1: by specifying '1)' you're removing it from the domain, meaning that it isn't even a point on the interval to begin with; it's a contradiction. If instead you just had a 'topological segment' or 'topological interval' from 0 --- 0, or just 0 -- , then I would think it would be homeomorphic to a circle with one point 0 along its curve. A topological interval from 0 --- 1, or A --- B however would be its own thing. With this in mind, I want to say that I share your aforementioned criticism of the real number line and I think that 0, negative infinity, and positive infinity can each be considered to be such points of interest along a topological interval like A --- B --- C, but I find great peace of mind in the idea that infinity should be treated as a single point much like zero, where its sign is really just representative of the direction from which it's approached, this means that the real number line or the positive and negative integers (not strictly the real number line but if confusing I can clarify later) is homeomorphic to a circle with two distinct points of interest at 0 and infinity. In other words, we can treat it like the pseudo-interval A --- B --- A and hence map it to a circle. What this means is that homeomorphism is heavily dependent upon these 'points of interest', and is not so much dependent on the shape of the interval. To illustrate this point: suppose you had a unit circle with two points of interest on it A and B -- for clarity's sake, place them on opposite sides (I apologise for not having a diagram). If we characterise two transformations A and B on such a space that creates the new space with two points of interest on it AA and AB for example (like matrix operations) on the same unit circle. and we define a AA = A, AB = BA = B, and BB = A, then we see that a transformation of A creates the same space with the same positioning, and a transformation of B flips the points of interest. This, I feel, is really what defines negative and positive numbers. We can extend this notion to four points, A,B,C,D, define a similar matrix-like relationships and where AX=X, BB = A, AB = BA = B, DD = B, etc. ensuring that all possible operations collapse into one of the original 4 points, and there we have defined the complex numbers -- we can observe that multiplying by C or D 'rotates' it by 90 degrees. Again we can extend such a space to higher degrees of notable points and occupy a topological surface instead and define the quaternions, etc. I imagine doing so ad nausium could create an arbitrarily precise rotatable space, but I'm yet to find that out. Anyway, I just wanted to convey my thoughts on the matter; thank you for such great lectures

[@abstractUser]

@maximuspower2 Yes, they're the same. It's easy to show (i) If C is the complex product of A and B, then OC || AB, and the converse i.e. (ii) If OC || AB, then C must represent the complex product of A and B, which should complete the proof of equivalence between the two approaches. However, it's interesting that the professor wants to avoid 'irrationalities' while sitting on one i.e the circle that 'revolves' around pi. A pi-less 'perspective' on circle, though an illusion, is interesting.

[@arekkrolak6320]

circle algebra - to do A*A technically you don't need this assumption with tangent, you can set any point B and use the rule (A*A)*B = A*(A*B) :)

[@TonyChan1986]

i think you did point that Circle -[1 point] is meromorphic to line A1 in the previous video

[@maximuspower2]

@abstractUser Yes I already said that they are the same. However, you probably shouldn't judge his views on rationals unless you fully heard his reasoning (or any other ideas anyone has).

[@monoman4083]

intetesting..

[@davesmith7528]

Since I, personally, had trouble making the connection between associativity and Pascal's Theorem, here is the corollary:- Proof that (AxB)xC = Ax(BxC) In the (possibly) intersecting hexagon A, B, C, AxB, O, BxC, A we have 1. AB//O(AxB) by construction 2. BC//O(BxC) by construction and since 2 pairs of // lines exist then 3. (AxB)C//A(BxC) the third pair must also be // by Pascal’s Theorem hence O(AxB)xC//OAx(BxC) by construction. But both these lines pass through O So they must be coincident lines, hence the points (AxB)xC and Ax(BxC) must also be coincident on the circle. Hence (AxB)xC=Ax(BxC) QED

[@ohault]

In the given (A*B)*C=(A*B)*C example, the group structure does not look like to be convex, but then, what are conditions to have a convex group structure ?

[@njwildberger]

Olivier Hault There is no notion of convex group structure that I know of.

[@KaewPeerasak]

very very usuful for a beginer as me

[@TheSara90]

the natural question from 43:08 is provable without pascal theorem;) it only requires math from the 6th grade, no need for points at infinity:) also aren't there suppose to be 2 lines at infinity, for this particular case??:) (43:54)

[@mathematics666]

You are a wonderful professor Excuse me, in fact I have two questions 1. What is the name of Topologic property that we used for comparison between the circl and line (the worm path). 2. in any way be a space of linear lines homeomorfic with circle because the line does notthe same as point topologicly

[@arkapointer]

Hani Mechri it is called contractibility. R is contractible but S1 is not. Note however a unit disk is contractible.

[@brendawilliams8062]

They in in shrinkable is the question.

[@horizon586]

the Indian high school teacher thing got me.

[@gorporpio]

Depends on what you mean by the meaning of is, I mean same. Bill Clinton?